SOLUTION: Consider the Function {{{f(x)=x^2+6x-2}}} Part A. Find h, the x-coordinate of the vertex of this parabola (show work) Part B. Substitute the two integers immediately to the ri

Algebra ->  Rational-functions -> SOLUTION: Consider the Function {{{f(x)=x^2+6x-2}}} Part A. Find h, the x-coordinate of the vertex of this parabola (show work) Part B. Substitute the two integers immediately to the ri      Log On


   

Question 148160: Consider the Function f%28x%29=x%5E2%2B6x-2 Part A. Find h, the x-coordinate of the vertex of this parabola (show work) Part B. Substitute the two integers immediately to the right of the h into the function to find the corresponding y values. Fill in a table and make sure your x-values are in increasing order in your table.
Answer by nabla(475) About Me  (Show Source):
You can put this solution on YOUR website!
A. There really isn't much work to show. You are applying the simple rule:
x=-b/(2a) is the vertex of a parabola in form f(x)=ax^2+bx+c. I will derive this for you after we apply the rule.
For f(x)=x^2+6x-2,
x=-6/2=-3 will be the x-coordinate of the vertex.
Derivation:
consider the derivative of the function, f'(x).
f'(x)=2x+6. For a quadratic equation, wherever the derivative is zero, we shall have a minimum or maximum, which by definition follows as the vertex.
So 0=2x+6, or x=-3. This is obviously the same answer we got from the rule, because it uses the same method:
f(x)=ax^2+bx+c
f'(x)=2ax+b
0=2ax+b
x=-b/(2a).
B. I'm not entirely certain what is wanted... f(-2) and f(-1)???
Just plug in the values into the equation and find the values. You can make a simple table likewise.